Let $f \in \Bbb Z[x]$ such that there exist distinct $a$, $b$, $c$ and $f(a)=f(b)=f(c)=1$. Prove that there is no $d \in \Bbb Z$ such that $f(d)=0$

You were in the right path, the idea is that if you have three distinct numbers in $\mathbb{Z}$ one must be different from $\pm1$.

Take $g(x)$ as you have defined and assume there is $d$ such that $f(d)=0$, then

$$(d-a)(d-b)(d-c)\tilde{g}(d)=-1$$

A contradiction since $d-a,d-b,d-c$ are distinct.

Your start is good.

So, we can write $f(x) =u(x)\cdot(x-a)(x-b)(x-c) + 1$ with some polynomial $u\in\Bbb Z[x]$.

Then, for any distinct integer $d$, we have that at least one of $|d-a|,\ |d-b|,\ |d-c|$ is $\ge 2$, and thus $u(d)\cdot(d-a)(d-b)(d-c)$ can't ever be $-1$.